$$6x \cdot \frac{x+4}{(x+7)(x+6)} = 0$$

$$ \begin{matrix}x_1 = 0 & x_2 = -4 & x_3 = -6 \end{matrix} $$

In order to solve $ \color{blue}{ 6x^{4}+96x^{3}+504x^{2}+864x = 0 } $, first we need to factor our $ x $.

$$ 6x^{4}+96x^{3}+504x^{2}+864x = x \left( 6x^{3}+96x^{2}+504x+864 \right) $$

$ x = 0 $ is a root of multiplicity $ 1 $.

The remaining roots can be found by solving equation $ 6x^{3}+96x^{2}+504x+864 = 0$.

$ \color{blue}{ 6x^{3}+96x^{2}+504x+864 } $ is a polynomial of degree 3. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $ \dfrac{p}{q} $, where p is a factor of the trailing constant and q is a factor of the leading coefficient.

The factors of the leading coefficient ( 6 ) are 1 2 3 6 .The factors of the constant term (864) are 1 2 3 4 6 8 9 12 16 18 24 27 32 36 48 54 72 96 108 144 216 288 432 864 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 1 }{ 6 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 3 } , ~ \pm \frac{ 2 }{ 6 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 3 } , ~ \pm \frac{ 3 }{ 6 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 4 }{ 3 } , ~ \pm \frac{ 4 }{ 6 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 6 }{ 3 } , ~ \pm \frac{ 6 }{ 6 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 2 } , ~ \pm \frac{ 8 }{ 3 } , ~ \pm \frac{ 8 }{ 6 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 2 } , ~ \pm \frac{ 9 }{ 3 } , ~ \pm \frac{ 9 }{ 6 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 2 } , ~ \pm \frac{ 12 }{ 3 } , ~ \pm \frac{ 12 }{ 6 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 16 }{ 2 } , ~ \pm \frac{ 16 }{ 3 } , ~ \pm \frac{ 16 }{ 6 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 18 }{ 2 } , ~ \pm \frac{ 18 }{ 3 } , ~ \pm \frac{ 18 }{ 6 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 24 }{ 2 } , ~ \pm \frac{ 24 }{ 3 } , ~ \pm \frac{ 24 }{ 6 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 27 }{ 2 } , ~ \pm \frac{ 27 }{ 3 } , ~ \pm \frac{ 27 }{ 6 } , ~ \pm \frac{ 32 }{ 1 } , ~ \pm \frac{ 32 }{ 2 } , ~ \pm \frac{ 32 }{ 3 } , ~ \pm \frac{ 32 }{ 6 } , ~ \pm \frac{ 36 }{ 1 } , ~ \pm \frac{ 36 }{ 2 } , ~ \pm \frac{ 36 }{ 3 } , ~ \pm \frac{ 36 }{ 6 } , ~ \pm \frac{ 48 }{ 1 } , ~ \pm \frac{ 48 }{ 2 } , ~ \pm \frac{ 48 }{ 3 } , ~ \pm \frac{ 48 }{ 6 } , ~ \pm \frac{ 54 }{ 1 } , ~ \pm \frac{ 54 }{ 2 } , ~ \pm \frac{ 54 }{ 3 } , ~ \pm \frac{ 54 }{ 6 } , ~ \pm \frac{ 72 }{ 1 } , ~ \pm \frac{ 72 }{ 2 } , ~ \pm \frac{ 72 }{ 3 } , ~ \pm \frac{ 72 }{ 6 } , ~ \pm \frac{ 96 }{ 1 } , ~ \pm \frac{ 96 }{ 2 } , ~ \pm \frac{ 96 }{ 3 } , ~ \pm \frac{ 96 }{ 6 } , ~ \pm \frac{ 108 }{ 1 } , ~ \pm \frac{ 108 }{ 2 } , ~ \pm \frac{ 108 }{ 3 } , ~ \pm \frac{ 108 }{ 6 } , ~ \pm \frac{ 144 }{ 1 } , ~ \pm \frac{ 144 }{ 2 } , ~ \pm \frac{ 144 }{ 3 } , ~ \pm \frac{ 144 }{ 6 } , ~ \pm \frac{ 216 }{ 1 } , ~ \pm \frac{ 216 }{ 2 } , ~ \pm \frac{ 216 }{ 3 } , ~ \pm \frac{ 216 }{ 6 } , ~ \pm \frac{ 288 }{ 1 } , ~ \pm \frac{ 288 }{ 2 } , ~ \pm \frac{ 288 }{ 3 } , ~ \pm \frac{ 288 }{ 6 } , ~ \pm \frac{ 432 }{ 1 } , ~ \pm \frac{ 432 }{ 2 } , ~ \pm \frac{ 432 }{ 3 } , ~ \pm \frac{ 432 }{ 6 } , ~ \pm \frac{ 864 }{ 1 } , ~ \pm \frac{ 864 }{ 2 } , ~ \pm \frac{ 864 }{ 3 } , ~ \pm \frac{ 864 }{ 6 } ~ $$

Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial $ P(x) $, we obtain $ P(-4) = 0 $.

To find remaining zeros we use Factor Theorem. This theorem states that if $\frac{p}{q}$ is root of the polynomial then this polynomial can be divided with $ \color{blue}{q x - p} $. In this example:

Divide $ P(x) $ with $ \color{blue}{x + 4} $

$$ \frac{ 6x^{3}+96x^{2}+504x+864 }{ \color{blue}{ x + 4 } } = 6x^{2}+72x+216 $$

Polynomial $ 6x^{2}+72x+216 $ can be used to find the remaining roots.

$ \color{blue}{ 6x^{2}+72x+216 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.

Equations Solver